## Monday, 10 April 2017

### Path Integral formulation - Part - 6 - Aharanov Bohm effect

In classical Electrodynamics, the Vector and Scalar potential is introduced as a pure mathematical tool to express the complex vector field quantities in its simplest form. But, it is always possible to explain any classical Electrodynamics phenomena entirely in terms of field quantities (Electric and Magnetic fields) alone.

Since, Maxwell's equations are entirely defined in terms of Electric and Magnetic field, even without the concept of vector and scalar potential, classical electrodynamics works just fine.  The field quantities are completely invariant under the change in vector and scalar potentials in a specific way. This property is popularly known as Gauge Invariance and the freedom to gauge transform the potentials is known as gauge freedom.

Using this freedom, one can chose for a given value of the field quantity, many different values of potential. And all the physical results remains the same under the gauge transformation, where different type of gauges are used for different problems.

Unlike the Maxwell's equations in classical electrodynamics, in Quantum domain, the fundamental Schr\"{o}dinger equation itself starts from the concept of Hamiltonian which is intrinsically based on the concept of vector and scalar potentials. So, when it is used the gauge transformation in Schr\"{o}dinger's equation, it is seen that everything remains the same except for the phase of the wave function that gets altered by the presence of vector or scalar potential.

Though it look like the potentials change the structure of the Quantum wave function, from the postulates of Quantum Mechanics, it is a well known fact that if a wave function is multiplied by any constant phase factor, it doesn't make any physical change in the probability distribution values. The probability of finding the particle at any position is the same independent of the value of the phase factor.

Thus, it is concluded that all the Quantum Mechanical results remains the same with respect to the gauge transformations of potentials and the phase of a Quantum wave function can never be experimentally measured.

It was until Aharanov and Bohm, who first showed that there is an experimental way to physically measure the change in the phase of a quantum wave function by measuring a shift in the interference pattern produced by electrons when it is placed a magnetic field nearby to the electron's trajectory using a long solenoid.

In this, the Double slit experiment is considered with a solenoid placed perpendicular to motional plane of the electrons from either side of the slits. Initially the solenoid is switched off and the normal interference pattern is observed on the screen.

Then the same experiment was repeated again by switching on the solenoid except now, the electrons gain a phase difference due to the presence of vector potential produced by the magnetic field inside the solenoid.
And a new interference pattern that is shifted by a small amount compared with the old interference pattern is observed on the screen.
It is to be noted that the electrons are only affected by the vector potential and not by the magnetic field produced by the solenoid, because the field outside the solenoid is zero.

Thus, the observation of this new interference pattern will serve as an experimental proof for the physical reality of the phase of the Quantum Wave function and so the physical reality of the fundamental vector and scalar potentials.

Though the original mathematics was worked out by Aharanov and Bohm using Schrodinger picture, it has a much simpler form in path Integral formulation.

$K(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L({x}(t), \dot{x}(t),t)}\,\tag{5.1}$
Since it is considered the motion of non-relativistic charged particle in an Electro-magnetic field, the Lagrangian is written as,
$L = L_0 + \frac{q}{c} A_i\frac{dx^i}{dt}\,\tag{5.2}$
In this set up, the scalar potential is not considered, so $A_0 = 0$. It is used $L_0$ for the Lagrangian of a free particle. The path integral is now,
$K(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0} e^{\frac{iq}{\hbar{c}}\int_{t_i}^{t_f}dt\,A_i\frac{dx^i}{dt}}\,\tag{5.3}$
The integral over the vector potential becomes a line integral over space
$K(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0} e^{\frac{iq}{\hbar{c}}\int_{x(t_i)=x_i}^{x(t_f)=x_f}dx^i\,A_i}\,\tag{5.4}$
With this integral, one can directly proceed to the experiment. In the first part, the solenoid is turned off in double slit set up which is equivalent to the regular double-slit experiment where the vector potential is zero. The particle behaves like a free particle.  To describe the interference pattern, one needs to take two path integrals on either side of the double slit.

It is calculated to be on the first side via path 1,
$K_1(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0}\,\tag{5.5}$
Similarly, on the other side via path 2,
$K_2(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0}\,\tag{5.6}$
These path integrals for a free particle which is already calculated to be,
$K_1(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}\,\tag{5.7}$
and
$K_2(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}\,\tag{5.8}$
Both of them can differ only by a phase difference. So, it is given a phase for each integral and written as,
$K_1(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}e^{\frac{i}{\hbar}\phi_1}\,\tag{5.9}$
and
$K_2(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}e^{\frac{i}{\hbar}\phi_2}\,\tag{5.10}$

Since, these path integrals are the real Schr\"{o}dinger wave function, they obey superposition rule. Thus, the total path integral is,
$K(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}e^{\frac{i}{\hbar}\phi_1} \left[ 1+ e^{\frac{i}{\hbar}(\phi_2-\phi_1)}\right]\,\tag{5.11}$
where the quantity
$\frac{\phi_2-\phi_1}{\hbar} = \Delta \,\tag{5.12}$
represents the phase difference between the two paths. The detector at this point $x_f$ will detect constructive interference if the phase difference between the waves is $2n\pi$  $$i.e.\,\,\,\,\Delta = 2n\pi$$ and destructive interference when the phase difference is $(2n+1)\pi$$i.e.\,\,\,\,\Delta=(2n+1)\pi$$ where$nis an integer in both the cases. To obtain the probability distribution, the modulus square of the probability amplitude is calculated, \begin{align*} K^*(x_f,&t_f;x_i,t_i)K(x_f,t_f;x_i,t_i) \\ &= \left\vert{K(x_f,t_f;x_i,t_i)}\right\vert^2 \\&= {\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[ 1+ e^{\frac{i}{\hbar}(\phi_2-\phi_1)}\right] \left[ 1+ e^{\frac{-i}{\hbar}(\phi_2-\phi_1)}\right]\,\,\tag{5.13} \\ &= {\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[ 2 + e^{\frac{i}{\hbar}(\phi_2-\phi_1)} + e^{\frac{-i}{\hbar}(\phi_2-\phi_1)}\right] \\ & = {\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[ 2 + 2cos\left(\frac{\phi_2-\phi_1}{\hbar}\right)\right] \\ & = {\frac{2m}{2i\pi\hbar{(t_f-t_i)}}}\left[ 2cos^2\left(\frac{\phi_2-\phi_1}{2\hbar} = 2 cos^2\left(\frac{\Delta}{2}\right)\right)\right] \end{align*} which gives finally, $\left\vert{K(x_f,t_f;x_i,t_i)}\right\vert^2 = {\frac{2m}{i\pi\hbar{(t_f-t_i)}}}cos^2\left(\frac{\Delta}{2}\right)\,\tag{5.14}$ which gives the normal sinusoidal interference pattern. Now, proceeding to the second part of the experiment, by switching on the solenoid and considering again the two path integrals. Except now, there is a non-zero vector potential around the solenoid. It is taken a very long solenoid compared to the distance between the slits such that the magnetic field all around the solenoid cancels off to zero except for the interior region of the solenoid. As far as the motion of the particle is concerned, the particle don't see any field as much as the same in the previous case. The only change comes in the phase of the propagator for the particle, which is written using (5.4), $$K(x_f,t_f; x_i,t_i) = \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0} e^{\frac{iq}{\hbar{c}}\int_{x(t_i)=x_i}^{x(t_f)x_f}dx^i\,A_i}$$ From the symmetry of the set up and considering symmetrical paths, one can assume with negligible error that, the line integral\int_i^fdx^i\,A_i$is independent of the path taken by the particle from the initial point$x_i$to the final point$x_f. So, taking this term out of the path integral and treat it as a phase coefficient much to the same of the previous case. $K(x_f,t_f; x_i,t_i) = e^{\frac{iq}{\hbar{c}}\int_{x(t_i)=x_i}^{x(t_f)x_f}dx^i\,A_i} \int_{{x}(t_i)={x_i}}^{{x}(t_f)={x_f}} \,d{x(t)} e^{\frac{i}{\hbar}\int_{t_i}^{t_f}dt\,L_0} \,\tag{5.15}$ And substituting for the integral using with corresponding phase factor for each path, for path 1, $K'_1(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}e^{\frac{i}{\hbar}\phi_1}e^{\frac{iq}{\hbar{c}}\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\,A_i} \,\tag{5.16}$ Similarly for path 2, $K'_2(x_f,t_f;x_i,t_i) =\sqrt{\frac{m}{2i\pi\hbar{(t_f-t_i)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}\right]}e^{\frac{i}{\hbar}\phi_2}e^{\frac{iq}{\hbar{c}}\int_{x_2(t_i)=x_i}^{x_2(t_f)=x_f}dx^i\,A_i} \,\tag{5.17}$ substituting for $$\frac{\phi_2-\phi_1}{\hbar} = \Delta$$ and combining the two integrals using superposition principle to derive the final probability amplitude, \begin{align*} K'&(x_f,t_f;x_i,t_i) ={\frac{m}{2i\pi\hbar{(t_f-t_i)}}}exp\left({\frac{i}{\hbar}\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}}\right)exp\left({\frac{i}{\hbar}\phi_1}\right)\\&exp\left({\frac{iq}{\hbar{c}}\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\,A_i}\right) \\&\left[1 + exp({i\Delta})exp\left({\frac{iq}{\hbar{c}}\left(\int_{x_2(t_i)=x_i}^{x_2(t_f)=x_f}dx^i\,A_i-\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\, A_i\right)}\right)\right] \,\,\,\tag{5.18} \end{align*} In the above equation, the last bracketed term of vector potential can be written as single closed integral form. $$\int_{x_2(t_i)=x_i}^{x_2(t_f)=x_f}dx^i\,A_i-\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\, A_i = -\left[\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\,A_i+\int_{x_2(t_i)=x_f}^{x_2(t_f)=x_i}dx^i\, A_i\right]$$ which is the negative of the closed integral fromx_i$to$x_f$and back again from$x_f$to$x_i$. So, $\int_{x_2(t_i)=x_i}^{x_2(t_f)=x_f}dx^i\,A_i-\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\, A_i = - \int_{closed\,loop} \,dx^i \, A_i\,\tag{5.19}$ Using, Stoke's theorem, the integral could be rewritten in terms of surface integral as, $\int_{closed\,loop} \,dx^i \, A_i = \int_{surface} \left(\nabla\times\vec{A}\right)\cdot\vec{dS}\,\tag{5.20}$ where the region of surface is just the cross section of the solenoid. But, the integrand is just the magnetic field$B\$, $$\vec{B} = \nabla\times\vec{A}$$ and the surface integral of the magnetic field is the magnetic flux, which in this case is the flux due to the solenoid alone.
$\int_{surface} \left(\nabla\times\vec{A}\right)\cdot\vec{dS} = \int_{surface}\vec{B}\cdot\vec{dS} = \Phi_B\,\tag{5.21}$
Using this, (5.18) becomes,
\begin{align*}
K'&(x_f,t_f;x_i,t_i) \\ &={\frac{m}{2i\pi\hbar{(t_f-t_i)}}}exp\left({\frac{i}{\hbar}\frac{m\,(x_f-x_i)^2}{2\,(t_f-t_i)}}\right) exp\left({\frac{i}{\hbar}\phi_1}\right) \\ & exp\left({\frac{iq}{\hbar{c}}\int_{x_1(t_i)=x_i}^{x_1(t_f)=x_f}dx^i\,A_i}\right) \left[1 + exp({i\Delta})exp\left({\frac{-iq\Phi_B}{\hbar{c}}}\right)\right]\,\tag{5.22}
\end{align*}
And the probability distribution given by the modulus square of this amplitude,

\begin{align*}
K'^*(x_f,t_f;x_i,t_i)&K'(x_f,t_f;x_i,t_i) \\&= \left\vert{K'(x_f,t_f;x_i,t_i)}\right\vert^2 \\&= {\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[1 + exp({i\Delta})exp\left({\frac{-iq\Phi_B}{\hbar{c}}}\right)\right] \\ & \left[1 + exp({-i\Delta})exp\left({\frac{iq\Phi_B}{\hbar{c}}}\right)\right]\,\,\tag{5.23}
\end{align*}

With the same arithmetic as previous,
\begin{align*}
&={\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[2 + exp({i\Delta})exp\left({\frac{-iq\Phi_B}{\hbar{c}}}\right) + exp({-i\Delta})exp\left({\frac{iq\Phi_B}{\hbar{c}}}\right)\right] \\& = {\frac{m}{2i\pi\hbar{(t_f-t_i)}}}\left[2 + 2cos\left(\left(\Delta-\frac{q\Phi_B}{\hbar{c}}\right)\right)\right]
\\& ={\frac{2m}{i\pi\hbar{(t_f-t_i)}}} cos^2\left(\frac{\left(\Delta-\frac{q\Phi_B}{\hbar{c}}\right)}{2}\right)\,\tag{5.24}
\end{align*}
Comparing our results with (5.14) it is observed that the interference pattern is shifted by the amount $$\frac{q\Phi_B}{\hbar{c}}$$. This shifting pattern is the effect of the solenoid on the charged particles. Even though, this shift is a very small amount and the experiment to observe this phenomena is a much complicated one, the result is perfectly valid. In 1986, this shift in the interference pattern is exactly observed by Akira Tonumara and his colleagues using the flux quantization in superconductors.

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